Is this understanding of a filtration correct?

Question

I am reading an article with some arguments about stochastic processes expressed in terms of the filtration of a vector of time-indexed random variables. I've gone through some lecture notes and videos I found online to give myself a crash course, and just wanted to clarify my understanding is correct. The parts I'm most unsure about are marked in bold.

1. Let $\Omega$ be the set of ALL possible outcomes that could ever be observed.
2. $\mathcal{F}$ is the largest $\sigma$-field generated by $\Omega$, denoted $2^\Omega$.
3. There is a strictly ordered set $T$ containing time indices.
4. We now have a sequence of random variables $X_t$ indexed by $t \in T$. The sequence of these time-indexed random variables, $X_1, X_2, ...$ is a stochastic process.
5. Each random variable is a mapping $X_t : \Omega \rightarrow \mathbb{R}$. In other words, each random variable is a function $x_t = X_t(\omega) s.t. x_t \in \mathbb{R}$. So each $X_t$ maps every possible outcome $\omega$ to a value in $\mathbb{R}$
6. $P$ is a mapping $P:\Omega \rightarrow [0,1]$ such that $\sum_{i=1}^{\infty}A_i = 1$ for any sequence of disjoint subsets $A_i$ where $\bigcup_{i=0}^{\infty}A_i = \Omega$
7. The triple ($\Omega, \mathcal{F}, P$) forms a probability space.
8. Now consider: we are interested in the relative frequencies of the sequences. What sort of structure would we like over the space?
9. A $\sigma$-algebra is a collection of sets with certain properties: closure under complement and countable union, and non-emptiness. Given some set $Y$, the smallest $\sigma$-algebra that contains $Y$ is called the $\sigma$-algebra generated by $Y$. It is also happens to be the intersection of all $\sigma$-algebras generated by $Y$. The properties of the $\sigma$-algebra make it measurable- we can define a measure such as a probability measure over it, and thus, indirectly, we are able to measure the original collection of interest.
10. The trivial $\sigma$-algebra generated by $\Omega$ is $\{\emptyset, \Omega\}$
11. Now imagine the process has 'started'. We first observe $X_1$, then $X_2$, and so on. We could be more precise and say that we observe some real values at timestep 1, timestep 2, and so on; each of these real values corresponds to the possible occurrence of a set of outcomes $\{\omega_a, \omega_b, ...\} \subseteq \Omega$
12. In order to measure the relative frequency of sequences of these random variables $\vec{X}_1^t$, we will need a $\sigma$-algebra over the collection of interest. In this case, we may well have a growing collection of events of interest, and thus a growing collection of $\sigma$-algebras. Note that the events of interest at this point are sequences of random variables.
13. As the sequences grow, so do the smallest $\sigma$-algebras that they generate. They grow in such a way that future $\sigma$-algebras are always supersets of previous $\sigma$-algebras.
14. A filtration is such a sequence of $\sigma$-algebras $\mathcal{F_1} \subseteq \mathcal{F_2} ... \subseteq \mathcal{F}$
15. Each sequence ${X_{1}}, X_1X_2, X_1X_2X_3...$ is measurable by the corresponding $\sigma$-algebras $\mathcal{F_1}, \mathcal{F_2},...$ Now these are the $\sigma$-algebras of the sequence random variables $\vec{X}_{1}^{t}$, so they are said to be adapted to $\vec{X}_{1}^{t}$, or $\vec{X}_{1}^{t}$-measurable.
16. If you had some other random variable $Y_t$ that is adapted to $\vec{X}_{1}^{t}$, would that imply that $Y_t$ must be a bijective function on the $\sigma$-algebra generated by $\vec{X}_{1}^{t}$?